洲阁筛

洲阁筛真是个不错的暴力啊。。

简单的写了个求1~n质数个数

-O2  N=1e11要2.5s

#pragma GCC optimize(2)
#include <bits/stdc++.h>
using namespace std;
#define LL long long
int L0[1000005],L1[1000005],sn,m,P[1000005],x,ga[1000005],gb[1000005];
LL n,a[1000005],b[1000005];
int main(){
    n=100000000000;
//    scanf("%lld",&n);
    sn=sqrt(n);
    for (int i=2;i<=sn;++i){
        if (!P[i]) P[++m]=i;
        for (int j=1;j<=m;++j){
            x=P[j]*i;
            if (x>sn) break;
            P[x]=1;
            if (i%P[j]==0) break;
        }
    }
    for (int i=1;i<=sn;++i) a[i]=i,b[i]=n/i;
    for (int i=1,j=1;i<=sn;++i){
        while (j<=m&&P[j]*P[j]<=i) ++j;
        L0[i]=j;
    }
    for (int i=sn,j=L0[sn];i;--i){
        while (j<=m&&P[j]*P[j]<=b[i]) ++j;
        L1[i]=j;
    }
    for (int i=1;i<=m;++i){
        for (int j=1;j<=sn&&i<L1[j];++j){
            LL y=j*P[i]; gb[j]=i;
            if (y<=sn) b[j]-=b[y]+gb[y]+1-i;
            else b[j]-=a[n/y]+ga[n/y]+1-i;
        }
        for (int j=sn;j&&i<L0[j];--j){
            LL y=j/P[i]; ga[j]=i;
            a[j]-=a[y]+ga[y]+1-i;
        }
    }
    printf("%lld\n",b[1]+m-1);
    return 0;
}

 

外加一个模板题 http://www.spoj.com/problems/DIVCNT3/

代码：

带上求f的那一部分，N=1e11要12s多

不开O2可能是要死的吧。。

#pragma GCC optimize(2)
#include <bits/stdc++.h>
using namespace std;
#define LL long long
int T,W[340005],L0[340005],L1[340005],L2[340005],sn,m,P[340005],x,ga[340005],gb[340005];
LL n,a[340005],b[340005],F[340005];
void hh(){
    scanf("%lld",&n);
    sn=sqrt(n); F[1]=1; m=0;
    for (int i=2;i<=sn;++i){
        if (!P[i]) P[++m]=i,F[i]=4,W[i]=1;
        for (int j=1;j<=m;++j){
            x=P[j]*i;
            if (x>sn) break;
            P[x]=1;
            if (i%P[j]==0){
                W[x]=W[i]+1;
                F[x]=F[i]/(3*W[i]+1)*(3*W[x]+1);
                break;
            }else{
                W[x]=1; F[x]=F[i]*4;
            }
        }
    }
    for (int i=1;i<=sn;++i)
        a[i]=i,b[i]=n/i,ga[i]=gb[i]=0;
    for (int i=1,j=1;i<=sn;++i){
        while (j<=m&&P[j]*P[j]<=i) ++j;
        L0[i]=j;
    }
    for (int i=sn,j=L0[sn];i;--i){
        while (j<=m&&1ll*P[j]*P[j]<=b[i]) ++j;
        L1[i]=j;
    }
    for (int i=1,j=1;i<=sn;++i){
        while (j<=m&&P[j]<=i) ++j;
        L2[i]=j;
    }
    for (int i=1;i<=m;++i){
        for (int j=1;j<=sn&&i<L1[j];++j){
            LL y=j*P[i]; gb[j]=i;
            if (y<=sn) b[j]-=b[y]+gb[y]+1-i;
            else b[j]-=a[n/y]+ga[n/y]+1-i;
        }
        for (int j=sn;j&&i<L0[j];--j){
            int y=j/P[i]; ga[j]=i;
            a[j]-=a[y]+ga[y]+1-i;
        }
    }
    LL ans=0;
    for (int i=1;i<=sn;++i)
        ans+=F[i]*((b[i]-m+gb[i]-1)*4+1);
    for (int i=1;i<=sn;++i)
        a[i]=b[i]=1,ga[i]=gb[i]=m+1;
    for (int i=m;i;--i){
        for (int j=1;j<=sn&&i<L1[j];++j){
            LL y=n/j;
            if (gb[j]!=i+1) b[j]=(m-i)*4+1;
            y/=P[i]; gb[j]=i;
            for (int c=1;y;++c,y/=P[i])
            if (y<=sn)
                if (ga[y]==i+1) b[j]+=a[y]*(3*c+1);
                else b[j]+=(max(0,L2[y]-i-1)*4+1)*(3*c+1);
            else
                if (gb[n/y]==i+1) b[j]+=b[n/y]*(3*c+1);
                else b[j]+=((m-i)*4+1)*(3*c+1);
        }
        for (int j=sn;j&&i<L0[j];--j){
            int y=j;
            if (ga[j]!=i+1) a[j]=max(0,L2[y]-i-1)*4+1;
            y/=P[i]; ga[j]=i;
            for (int c=1;y;++c,y/=P[i])
            if (ga[y]==i+1) a[j]+=a[y]*(3*c+1);
            else a[j]+=(max(0,L2[y]-i-1)*4+1)*(3*c+1);
        }
    }
    ans+=b[1]-(ga[sn]==1?a[sn]:m*4+1);
    for (int i=1;i<=sn;++i) P[i]=0;
    printf("%lld\n",ans);
}
int main(){
    scanf("%d",&T);
    while (T--) hh();
    return 0;
}